Solid–Vapour Equilibrium

Solid–Vapour Equilibrium

When a solid sublimes, it passes directly into the vapour phase without melting. If you place solid iodine in a closed vessel, you will soon see violet vapour filling the space above the solid. The colour deepens with time, but after a while it stops changing — the intensity becomes constant. At that point, the system has reached equilibrium.

The forward process is sublimation: solid iodine turns into iodine vapour. The reverse process is condensation: iodine vapour deposits back into solid iodine. At equilibrium, both processes occur at the same rate, so the amount of vapour (and hence the colour intensity) remains steady.

The equilibrium is written as:

Other common examples of solid–vapour equilibrium include:

• Camphor (solid) $\rightleftharpoons$ Camphor (vapour)
• $\text{NH}_4\text{Cl}$ (solid) $\rightleftharpoons$ $\text{NH}_4\text{Cl}$ (vapour)

Key Idea: Vapour Pressure of a Solid

At a fixed temperature, the vapour pressure of a solid in equilibrium with its vapour is constant. This is analogous to the vapour pressure of a liquid. For example, solid iodine at room temperature has a measurable vapour pressure — that is what fills the vessel with violet vapour. The equilibrium vapour pressure depends only on the temperature, not on the amount of solid present (as long as some solid remains).

Properties of Solid–Vapour Equilibrium

Several important properties follow from the nature of this equilibrium. Each is stated and then explained fully.

Property 1: The equilibrium is dynamic

At equilibrium, sublimation and condensation continue to occur at equal rates. The system is not static — molecules are constantly leaving the solid surface and returning to it. The net change is zero, but the microscopic exchange is ongoing.

Proof/Derivation:

Consider a closed vessel containing solid iodine. Initially, only sublimation occurs, so the vapour concentration rises. As vapour builds up, the rate of condensation (vapour molecules striking the solid and sticking) increases. Eventually, the rate of condensation equals the rate of sublimation. At that point, the number of molecules leaving the solid per second equals the number returning per second. The vapour pressure becomes constant. This is a dynamic equilibrium — the forward and reverse rates are equal, not zero.

Property 2: The vapour pressure is constant at a given temperature

For a fixed temperature, the equilibrium vapour pressure of a solid is a fixed value, independent of the amount of solid or the volume of the container (as long as some solid is present).

Proof/Derivation:

The equilibrium condition for the reaction $\text{Solid} \rightleftharpoons \text{Vapour}$ is given by the equality of chemical potentials:

For a pure solid, $\mu_{\text{solid}}$ depends only on temperature and pressure (very weakly on pressure for solids). For the vapour, assuming ideal behaviour:

where $P$ is the vapour pressure. At equilibrium, $\mu_{\text{solid}}$ is fixed at a given $T$, so $\ln P$ must be fixed. Hence $P$ is constant. Changing the amount of solid or the container volume does not change $\mu_{\text{solid}}$ (as long as solid is present), so $P$ remains the same.

Property 3: The vapour pressure increases with temperature

As temperature rises, more molecules have enough energy to escape the solid lattice, so the equilibrium vapour pressure increases.

Proof/Derivation:

This follows from the Clausius–Clapeyron equation, which relates vapour pressure to temperature:

where $\Delta H_{\text{sub}}$ is the enthalpy of sublimation (positive, since sublimation is endothermic), $R$ is the gas constant, and $C$ is a constant. Differentiating with respect to $T$:

Since $\Delta H_{\text{sub}} > 0$, the derivative is positive — so $P$ increases with $T$. The higher the temperature, the greater the vapour pressure.

Property 4: The equilibrium constant $K$ equals the vapour pressure

For the reaction $\text{Solid} \rightleftharpoons \text{Vapour}$, the equilibrium constant expression is:

But the concentration of a pure solid is constant (its density divided by molar mass) and is incorporated into the equilibrium constant. In terms of partial pressures (for ideal gases), the equilibrium constant $K_p$ is simply the partial pressure of the vapour at equilibrium:

Proof/Derivation:

The standard equilibrium constant expression for a heterogeneous reaction involving a pure solid and a gas is:

where $P^\circ$ is the standard pressure (1 bar or 1 atm, depending on convention). Often, $P^\circ$ is omitted for simplicity, and we write $K_p = P_{\text{vapour}}$. The solid does not appear because its activity is 1. So the equilibrium constant is numerically equal to the vapour pressure of the solid (in the chosen pressure units). …